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Uniformly cordial graphs
Discrete Mathematics 306 (2006) 726 – 737 www.elsevier.com/locate/disc
Uniformly cordial graphs Gary Chartranda , Sin-Min Leeb , Ping Zhanga,1 a Department of Mathematics, Western Michigan University, Kalamazoo, MI 49008, USA b Department of Computer Sciences, San Jose State University, San Jose, CA 95192, USA
Received 24 May 2002; received in revised form 7 November 2004; accepted 7 November 2005 Available online 31 March 2006 To the Memory of W. T. Tutte
Abstract Let G be a graph with vertex set V (G) and edge set E(G). A labeling f : V (G) → {0, 1} induces an edge labeling f ∗ : E(G) → {0, 1}, defined by f ∗ (xy) = |f (x) − f (y)| for each edge xy ∈ E(G). For i ∈ {0, 1}, let ni (f ) = |{v ∈ V (G) : f (v) = i}| and mi (f ) = |{e ∈ E(G) : f ∗ (e) = i}|. Let c(f ) = |m0 (f ) − m1 (f )|. A labeling f of a graph G is called friendly if |n0 (f ) − n1 (f )| 1. A cordial labeling of G is a friendly labeling f for which c(f ) 1. A graph G is a uniformly cordial graph if every friendly labeling of G is cordial. It is shown that a connected graph G of order n 2 is uniformly cordial if and only if n = 3 and G = K3 , or n is even and G = K1,n−1 . © 2006 Elsevier B.V. All rights reserved. Keywords: Friendly labeling; Cordial labeling; Uniformly cordial graph
1. Introduction We refer to the book [4] for graph theory notation and terminology not described in this paper. Let G be a graph with vertex set V (G) and edge set E(G). A labeling f : V (G) → {0, 1} induces an edge labeling f ∗ : E(G) → {0, 1}, defined by f ∗ (xy) = |f (x) − f (y)| for each edge xy ∈ E(G). For i ∈ {0, 1}, let ni (f ) = |{v ∈ V (G) : f (v) = i}| and
mi (f ) = |{e ∈ E(G) : f ∗ (e) = i}|.
Let c(f ) = |m0 (f ) − m1 (f )|. A labeling f of a graph G is called friendly if |n1 (f ) − n0 (f )| 1. A cordial labeling is a friendly labeling f for which c(f )1. Note that interchanging the vertex labels 0 and 1 in a cordial labeling results in a new cordial labeling of G. A cordial graph is a graph that admits a cordial labeling. Cordial labelings of graphs were introduced by Cahit [2], who showed that (1) every tree is cordial, (2) Kn is cordial if and only if n 3, (3) Kr,s is cordial for all r and s, (4) the wheel Wn is cordial if and only if n ≡ 3 (mod 4), (5) Cn is cordial if and only if n = 2 (mod 4), and (6) an Eulerian graph is not cordial if its size is congruent to 2 modulo 4. Benson and Lee [1] described a large class of regular cordial windmill graphs. Du [5] investigated cordial complete k-partite graphs. Kuo et al. [12] determined all m and n for which mK n is cordial. Lee, et al. [14] exhibited some cordial graphs. Generalized Petersen graphs that are cordial are characterized in [7]. Ho et al. [8] investigated the construction 1 Research supported in part by a Western Michigan University Faculty Research and Creative Activities Fund.
E-mail address: [email protected] (P. Zhang). 0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2005.11.025
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of cordial graphs using Cartesian products and compositions of graphs. Shee and Ho [17] determined the cordiality of (n) Cm , the one-point union of n copies of Cm . Several constructions of cordial graphs were proposed in [13,15–17,9–11]. Other results and open problems concerning cordial graphs are given in [3,6]. A graph G is defined to be uniformly cordial if every friendly labeling of G is cordial. That is, a graph G is uniformly cordial if whenever V (G) is partitioned at random into two subsets whose cardinalities are as equal as possible, then this induces a cordial labeling of G. For example, K3 is uniformly cordial. Since Kn , n4, is not cordial, it is not uniformly cordial either. Proposition 1.1. The complete graph Kn , n2, is a uniformly cordial graph if and only if n 3. 2. Some preliminary results We first determine all uniformly cordial complete bipartite graphs. Proposition 2.1. A complete bipartite graph Kr,s , where 1 r s, is uniformly cordial if and only if r = 1 and s is odd. Proof. Let G = Kr,s , where 1 r s. First, assume that r = 1 and s is odd. Then s = 2k − 1 for some integer k 1. Let f be a friendly labeling of G, where Vi is the set of vertices of G labeled i for i = 0, 1. Then |V0 | = |V1 | = k. Let u be the central vertex of G. Assume, without loss of generality, that u ∈ V0 . Then m0 (f ) = k − 1 and m1 (f ) = k. Since c(f ) = 1, it follows that f is cordial. Therefore, G is uniformly cordial. For the converse, assume that G = K1,s for any odd integer s. We consider two cases. Case 1: G is a star. Then r = 1 and s is even. Let s = 2k for some integer k 1. Let V (G) = {u, u1 , u2 , . . . , us }, where u is the central vertex of G. Define a labeling f of G by f (u) = f (ui ) = 0 for 1i k − 1, and f (ui ) = 1 for k i 2k. Since n0 (f ) = k and n1 (f ) = k + 1, it follows that f is a friendly labeling of G. On the other hand, m0 (f ) = k − 1 and m1 = k + 1. Thus c(f ) = 2 and so f is not cordial. Therefore, G is not uniformly cordial. Case 2: G is not a star. Then 2 r s. Let X and Y be the partite sets of G with |X| = r and |Y | = s. If r = s, then we define a friendly labeling f of G by assigning 0 to each vertex in X and 1 to each vertex in Y . Then m0 (f ) = 0 and m1 (f ) = r 2 . Since c(f ) = r 2 4, it follows that f is not cordial. Thus we may assume that 2 r < s. Partition V (G) into V0 and V1 such that V0 | − |V1 1 and X ⊆ V0 . Define a friendly labeling f of G by assigning i to every vertex in Vi for i = 0, 1. There are two cases. Subcase 2.1: r + s is even. Then r + s = 2k for some integer k 2. Thus |V0 | = |V1 | = k. Then m0 (f ) = r(k − r) and m1 (f ) = rk. Since c(f ) = r 2 4, it follows that f is not cordial. Subcase 2.2: r + s is odd. Then r + s = 2k + 1 for some integer k 2. Let |V0 | = k and |V1 | = k + 1. Then m0 (f ) = r(k − r) and m1 (f ) = r(k + 1). Since c(f ) = r 2 + r 6, it follows that f is not cordial. Therefore, G is not uniformly cordial in this case. It will be useful to know that certain classes of graphs are not uniformly cordial. Lemma 2.2. Let n3. If G is obtained from Kn−1 by adding a pendant edge, then G is not uniformly cordial. Proof. Suppose that G is obtained from Kn−1 by adding the pendant edge uv, where u ∈ V (Kn−1 ). We consider two cases. Case 1: n is odd. Then n = 2k + 1 for some integer k 1. Partition V (Kn−1 ) into V0 and V1 such that |V0 | = |V1 | = k and u ∈ V0 . Define a friendly labeling f of G by assigning i to every vertex in Vi for i = 0, 1 and assigning 1 to the k = k 2 − k and m1 (f ) = k 2 + 1. Since |m0 (f ) − m1 (f )| = k + 1 2, it follows that f vertex v. Then m0 (f ) = 2 2 is not cordial. Case 2: n is even. The graph G is not uniformly cordial for n = 4 and n = 6. Friendly labelings of G that are not cordial are shown in Fig. 1 for n = 4 and n = 6. For n 8, let n=2k, where then k 4. Now partition V (G) into V0 and V1 such that |V0 |=|V1 |=kandu, v ∈ V0 . Define k + +1=k 2 −2k +2 a friendly labeling f of G by assigning i to every vertex in Vi for i =0, 1. Then m0 (f )= k−1 2 2
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Fig. 1. Friendly labelings that are not cordial for n = 4, 6.
and m1 (f ) = (k − 1)k = k 2 − k. Since k 4, it follows that |m0 (f ) − m1 (f )| = k − 2 2 and so f is not cordial. Lemma 2.3. Let n 5. If G = Kn − M for some matching M in Kn , then G is not uniformly cordial. Proof. Let M = {e1 , e2 , . . . , e }, where ei = xi yi for 1 i . We consider two cases. Case 1: n is odd. Let n = 2k + 1, where then k 2. Partition V (G) into V0 and V1 such that (1) V0 | − |V1 = 1 and (2) xi and yi both belong to V0 or xi and yi both belong to V1 for eachi (1i ). Define a friendly labeling f of G by k + − = k 2 − and m1 (f ) = k(k + 1) = k 2 + k. assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = k+1 2 2 Since c(f ) = k + 3, it follows that f is not cordial. Case 2: n is even. Let n = 2k, where then k 3. There are two subcases. Subcase 2.1: k is odd and =k. Let k=2a+1. Partition V (G) into V0 and V1 such that V0 ={xi , yi : 1 i a}∪{x2a+1 } and V1 = {xi , yi : a + 1i 2a} ∪ {y2a+1 }. Then |V0 | = |V1 | = k. Define a friendly labeling f of G by assigning i to k every vertex in Vi for i =0, 1. Then m0 (f )=2 −(k −1)=(k −1)2 and m1 (f )=k 2 −1. Since c(f )=2(k −1) 4, 2 it follows that f is not cordial. Subcase 2.2: Subcase 2.1. does not occur. Partition V (G) into V0 and V1 such that (1) |V0 | = |V1 | = k and (2) xi ∗ and yi both belong to V0 or xi and yi both belong to V1 for each i (1 i). Define a friendly labeling f of G by k k assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ∗ ) = + − = k(k − 1) − and m1 (f ∗ ) = k 2 . Since 2 2 c(f ∗ ) = k + 4, it follows that f ∗ is not cordial. 3. Which graphs are randomly cordial? In an attempt to characterize all connected graphs that are uniformly cordial, we begin by showing that all connected graphs (of sufficiently large odd order) are not uniformly cordial. For a vertex v in a graph G, let N (v) denote the neighborhood of v and N [v] = N (v) ∪ {v}. 3.1. Graphs with odd order Theorem 3.1. No connected graph of odd order n 5 is a uniformly cordial graph. Proof. Assume, to the contrary, that there is a connected graph G of order n = 2k + 1 5 that is uniformly cordial. By Propositions 1.1 and 2.1 and Lemmas 2.2 and 2.3, we may assume that G is not the complete graph Kn , the star K1,n−1 , the graph Kn−1 with a pendant edge, or Kn − M for some matching M in G. Thus, there exists a vertex u in G such that 2 deg u = n − 3 = 2k − 2. Let N (u) = {u1 , u2 , . . . , u } and V (G) − N [u] = {v1 , v2 , . . . , v2k− }. Consider the sequence s : u, u1 , u2 , . . . , u , v1 , v2 , . . . , v2k− of the vertices of G. Partition V (G) into two subsets V0
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and V1 such that V0 consists of the first k + 1 vertices of s. Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. By assumption, f is a cordial labeling of G and so c(f ) = |m0 (f ) − m1 (f )| 1. Define a labeling f of G by f (u) = 1 and f (v) = f (v) for all v ∈ V (G) − {u}. Then n0 (f ) = k and n1 (f ) = k + 1, implying that f is a friendly labeling of G as well. We show that f is not a cordial labeling of G. There are two cases. Case 1: k. Then m0 (f ) = m0 (f ) − and m1 (f ) = m1 (f ) + . Thus m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) − 2. Since c(f )1 and 24, it follows that c(f ) 3. Case 2: k+1. Then m0 (f )=m0 (f )−k+(−k)=m0 (f )+−2k and m1 (f )=m1 (f )+k−(−k)=m1 (f )−+2k. Thus m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) + 2 − 4k = m0 (f ) − m1 (f ) − 2(2k − ). Since c(f )1 and 2k − 2, it follows that c(f ) 3. In either case, f is not a cordial labeling of G, which is a contradiction.
Corollary 3.2. Let G be a connected graph of odd order n3. Then G is a uniformly cordial graph if and only if G = K3 . In Proposition 2.1, we determined all uniformly cordial bipartite graphs. Theorem 3.1 can be used to show that, with the exception of K3 , there are no complete multipartite graphs that are not stars. Proposition 3.3. For k 3, no complete k-partite graph different from K3 is uniformly cordial. Proof. Assume, to the contrary, that there is a complete k-partite G with G = K3 that is uniformly cordial, where k 3. Let G = Kn1 ,n2 ,...,nk , where n1 n2 · · · nk and n = n1 + n2 + · · · + nk . By Theorem 3.1, n is even. Let n = 2t for some integer t 2 and let X1 , X2 , . . . , Xk be the partite sets of G such that |Xi | = ni for 1 i k. We consider two cases. Case 1: n1 t + 1. Partition V (G) into V0 and V1 such that |V0 | = |V1 | = t and V0 ⊆ X1 . Then V1 = V11 ∪ V12 such that V11 = X1 ∩ V1 and |V12 | = 2. Define a friendly labelingf by assigning i to every vertex in Vi for i = 0, 1. Note that m0 (f ) = (t − ) and m1 (f ) = t. Since c(f ) = 2 4, it follows that f is not cordial, which is a contradiction. Case 2: n1 t. There are two subcases. Subcase 2.1: n2 2. Partition V (G) into V0 and V1 such that (1) |V0 |=|V1 |=t and (2) X1 ⊂ V0 and X2 ⊆ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. By assumption, f is a cordial labeling of G and so c(f ) = |m0 (f ) − m1 (f )|1. Let u ∈ X1 and v ∈ X2 . Thus f (u) = 0 and f (v) = 1. Define a new friendly labeling f by f (u) = 1, f (v) = 0, and f (x) = f (x) for each x ∈ V (G) − {u, v}. Now m0 (f ) = m0 (f ) + (n1 − 1) + (n2 − 1). Therefore, m1 (f )=mi (f )−(n1 −1)−(n2 −1). It follows then that m1 (f )−m0 (f )=m1 (f )−m0 (f )−2(n1 +n2 −2). Since n1 + n2 − 2 2 and c(f )1, it follows that c(f ) 3 and so f is not cordial, a contradiction. Subcase 2.2: n2 = 1. Hence n1 t and ni = 1 for 2 i k. Since G is not a complete graph, n1 2. This implies that G = K n1 + Kn−n1 . Partition V (G) into V0 and V1 such that |V0 | = |V1 | = t and X 1 ⊆ V0 . Define a friendly labeling f ∗ of G by assigning i to each vertex in Vi for i = 0, 1. Then m0 (f ∗ ) = 2 2t − n21 and m1 (f ∗ ) = t 2 . Thus c(f ∗ ) = t + n21 3, implying that f ∗ is not cordial. Therefore, in either case, G is not uniformly cordial. 3.2. Graphs of even order with diameter at least 3 We now know that if there is any uniformly cordial graph G that is neither K3 nor a star, then G must have even order. We have seen in Proposition 1.1 that K3 is the only complete graph that is uniformly cordial. Since complete graphs are the only connected graphs with diameter 1, we assume that all connected graphs under consideration have even order and diameter at least 2. In this section, we show that there exists no uniformly cordial graph of even order with diameter 3 or more. In order to do this, we first show that there are certain properties that no uniformly cordial graphs of even order may possess.
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For two nonadjacent vertices u and v in a graph G, let N (u, v) = N (u) ∩ N (v), N ∗ (u, v) = N (u) − N (u, v), and If the two nonadjacent vertices u and v under discussion are clear, then we simply write N ∗ (u, v) = N ∗ (u) and N ∗ (v, u) = N ∗ (v). N ∗ (v, u) = N(v) − N (u, v).
Lemma 3.4. Let G be a connected graph of even order 2k 4. If G contains two nonadjacent vertices u and v such that (1) |N ∗ (u)| + |N ∗ (v)|2 and (2) |N ∗ (u)|k − 1 and |N ∗ (v)| k − 1, then G is not uniformly cordial. Proof. Let |N ∗ (u)| = a and |N ∗ (v)| = b. Partition V (G) into two subsets V0 and V1 with |V0 | = |V1 | = k such that {u} ∪ N ∗ (u) ⊆ V0 and {v} ∪ N ∗ (v) ⊆ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. By assumption, f is a cordial labeling of G and so c(f ) = |m0 (f ) − m1 (f )| 1. Define a new friendly labeling f by f (u) = 1, f (v) = 0, and f (x) = f (x) for all x ∈ V (G) − {u, v}. Since m0 (f ) = m0 (f ) − a − b
and
m1 (f ) = m1 (f ) + a + b,
it follows that m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) − 2(a + b). Since a + b 2 and c(f )1, it follows that c(f ) 3. Therefore, G is not uniformly cordial.
Lemma 3.5. Let G be a connected graph of even order 2k 4. If G contains two vertices u and v such that |N ∗ (u)| k and |N ∗ (u)| − |N ∗ (v)|2k − 4, then G is not uniformly cordial. Proof. Let |N ∗ (u)| = a and |N ∗ (v)| = b. Thus a k and a − b 2k − 4. Partition V (G) into two subsets V0 and V1 with |V0 | = |V1 | = k such that V0 consists of u and k − 1 vertices from N ∗ (u) − {v}. Then v ∈ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. By assumption, f is a cordial labeling of G and so c(f ) = |m0 (f ) − m1 (f )|1. Define a new friendly labeling f by f (u) = 1, f (v) = 0, and f (x) = f (x) for all x ∈ V (G) − {u, v}. If u and v are nonadjacent, then m0 (f ) = m0 (f ) − (k − 1) + [a − (k − 1)] − b, m1 (f ) = m1 (f ) + (k − 1) − [a − (k − 1)] + b. If u and v are adjacent, then m0 (f ) = m0 (f ) − (k − 1) + [(a − 1) − (k − 1)] − (b − 1), m1 (f ) = m1 (f ) + (k − 1) − [(a − 1) − (k − 1)] + (b − 1). In either case, m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) − 2[2(k − 1) − a + b]. Note that a − b 2k − 4 and so 2(k − 1) − a + b 2. Since c(f ) 1, it follows that c(f ) 3. Therefore, G is not uniformly cordial. Proposition 3.6. Let G be a connected graph of even order 2k 4. If G contains two nonadjacent vertices u and v such that (1) N (u, v) = ∅ or (2) N ∗ (u) = ∅ and N ∗ (v) = ∅, then G is not uniformly cordial. Proof. Let |N ∗ (u)| = a and |N ∗ (v)| = b. We consider two cases. Case 1: N (u, v) = ∅. Then N ∗ (u) = N (u) and N ∗ (v) = N (v). Since (1) G is connected, (2) u and v are nonadjacent, and (3) N (u, v) = ∅, it follows that 1 a 2k − 3
and
1b 2k − 3.
If a k − 1 and b k − 1, then G is not uniformly cordial by Lemma 3.4 since a + b 2. Thus, assume that one of a and b is at least k, say a k. Since a − b (2k − 3) − 1 = 2k − 4, it follows by Lemma 3.5 that G is not uniformly cordial.
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Case 2: N ∗ (u) = ∅ and N ∗ (v) = ∅. Thus, a, b 1. Since u and v are nonadjacent, it follows that 1 a 2k − 3 and 1b 2k − 3. An argument similar to one used in Case (1) shows that G is not uniformly cordial. The distance d(u, v) between two vertices u and v in a connected graph G is the length of a shortest u − v path in G. For a vertex u in a connected graph G, the eccentricity e(u) of u is the distance between u and a vertex of G that is farthest from u. The minimum eccentricity among the vertices of G is its radius rad G and the maximum eccentricity is its diameter diam G. Corollary 3.7. Let G be a connected graph of even order n4. If diam G3, then G is not uniformly cordial. Proof. Let u and v be two vertices of G such that d(u, v) 3. Thus u and v are nonadjacent and N (u, v) = ∅. It then follows by Proposition 3.6 that G is not uniformly cordial. 3.3. Graphs of even order with diameter 2 By Corollary 3.7 no connected graph of even order with diameter 3 or more is uniformly cordial. We now consider connected graphs of even order with diameter 2. We first show that there are certain properties that no uniformly cordial graph of even order having diameter 2 may possess. Some additional definitions and notation will be useful. For a graph G and a set S of vertices of G, let G be the complement of G and let S be the subgraph of G induced by S. For two vertex-disjoint graphs G and H , let G ∪ H be the union of G and H and let G + H be the join of G and H . Lemma 3.8. Let G be a connected graph of even order n = 2k 4 with diam G = 2. If G contains two nonadjacent vertices u and v such that |N ∗ (u)|k and N ∗ (v) = ∅, then G is not uniformly cordial. Proof. If |N ∗ (u)| 2k − 4, then G is not uniformly cordial by Lemma 3.5. Thus we may assume that |N ∗ (u)| 2k − 3. Since diam G = 2, it follows that N (u, v) = ∅, which implies that |N ∗ (u)| 2k − 3. Therefore, |N ∗ (u)| = 2k − 3 and |N(u, v)| = 1. Let N (u, v) = {w}. Again, since diam G = 2, the vertex w is adjacent to every vertex in N ∗ (u). Let H = N ∗ (u). We consider three cases. Case 1: H is complete. Then G is obtained from Kn−1 by adding a pendant edge vw. Since n4, it then follows by Lemma 2.2 that G is not uniformly cordial. Case 2: H is an empty graph. Then G is obtained from K1,1,2k−3 , where u and w are two vertices of eccentricity 1 in K1,1,2k−3 , by adding a new vertex v and the edge vw. If k = 2, then G is obtained by adding a pendant edge to K3 and so G is not uniformly cordial by Lemma 2.2. For k 3, partition V (G) into V0 and V1 with |V0 | = |V1 | = k such that u, v, w ∈ V0 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = 2 + 2(k − 3) = 2k − 4
and
m1 (f ) = 2k.
Since c(f ) = 4, it follows that G is not uniformly cordial. Case 3: H is neither complete nor empty. Then H contains three vertices x, y, and z such that xy ∈ / E(H ) and xz ∈ E(H ). Consider the two nonadjacent vertices v and x in G. Since z, u ∈ N ∗ (x, v) and v, w, x, y ∈ / N ∗ (x, v), it follows that 2 |N ∗ (x, v)| 2k − 4. There are two subcases. Subcase 3.1: |N ∗ (x, v)|k. Since |N ∗ (x, v)| 2k − 4, it follows by Lemma 3.5 that G is not uniformly cordial. Subcase 3.2: |N ∗ (x, v)|k − 1. Since |N ∗ (v, x)| = 0 k − 1 and |N ∗ (x, v)| + |N ∗ (v, x)| = |N ∗ (x, v)| 2, it follows by Lemma 3.4 that G is not uniformly cordial. We now have the following corollary.
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Corollary 3.9. Let G be a connected graph of even order n 4 with diam G=2. If G contains two nonadjacent vertices u and v with |N ∗ (u)| + |N ∗ (v)|2, then G is not uniformly cordial. Proof. Let n = 2k 4. We consider two cases. Case 1: |N ∗ (u)| k − 1 and |N ∗ (v)|k − 1. By Lemma 3.4, G is not uniformly cordial. Case 2: Exactly one of |N ∗ (u)| and |N ∗ (v)| is at least k, say |N ∗ (u)| k. There are two subcases. Subcase 2.1: |N ∗ (v)| = 0. By Lemma 3.8, G is not uniformly cordial. Subcase 2.2: |N ∗ (v)| 1. Since diam G = 2, it follows that N (u, v) = ∅ and so |N ∗ (u)| 2k − 4. By Lemma 3.5, G is not uniformly cordial. It remains to determine which connected graphs G of even order n 4 with diam G=2 such that |N ∗ (u)|+|N ∗ (v)| 1 for every two nonadjacent vertices u and v of G are uniformly cordial. We begin with the case where |N ∗ (u)| + |N ∗ (v)| is constant for every two nonadjacent vertices u and v of G. Lemma 3.10. Let G be a connected graph of even order n 4 with diam G = 2. If |N ∗ (u)| + |N ∗ (v)| = 1 for every two nonadjacent vertices u and v of G, then G = rP 3 ∪ sK 1 for some integers r 1 and s 0. Proof. Observe that if v1 and v2 are two nonadjacent vertices of G, then | deg v1 − deg v2 | = 1. Let u be a vertex of G with e(u) = 2. For i = 1, 2, let Ai be the set of vertices of G at distance i from u. Thus Ai = ∅ for i = 1, 2 and V (G) = {u} ∪ A1 ∪ A2 . Let v ∈ A2 . So uv ∈ / E(G). By hypotheses, |N ∗ (u)| + |N ∗ (v)| = 1 and so exactly one of N ∗ (u) ∗ and N (v) is empty. Without loss of generality, we assume that N ∗ (u) = ∅; otherwise, we interchange the roles of u and v. Let Fi = Ai for i = 1, 2. Claim 1. F2 = K2 . Proof of Claim 1. Since |N ∗ (u)| + |N ∗ (v)| = 1 and N ∗ (u) = ∅, there exists exactly one vertex, say w, in F2 that is adjacent to v in F2 . All vertices of G that are adjacent to u are necessarily adjacent to v as well, that is, u and v are mutually adjacent to the vertices in A1 . Likewise, u and w are not adjacent. Since v is adjacent to w but not to u, every vertex in A1 is adjacent to w as well (see Fig. 2). Since vw ∈ E(G) and v, w ∈ A2 , it follows that F2 contains K2 . Assume, to the contrary, F2 = K2 . Then there exists a vertex x ∈ A2 − {v, w} and vx ∈ / E(G). If x is adjacent to a vertex y ∈ A2 , then N (x) = {y} ∪ A1 and / E(G). So x is an isolated vertex in F2 . Since ux ∈ / E(G), it degG v = degG x; however, this is impossible since vx ∈ follows that |N ∗ (u)| + |N ∗ (x)| = 1. This implies that x is adjacent to all but one vertex y in A1 . Since (1) x is not
Fig. 2. A step in the proof of Claim 1.
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Fig. 3. A graph G = P3 ∪ 3K1 in 3.10 for r = 1 and s = 3.
adjacent to v and w and (2) y is adjacent to v and w, it follows that |N ∗ (x)| + |N ∗ (y)| 2, which is a contradiction. Therefore, Fi = A2 = K2 , as claimed. By Claim 1, we have G = (K1 ∪ K2 ) + F1 . If F1 is complete, then G = P3 ∪ (n − 3)K1 and we have the desired result. Thus we may assume that F1 is not complete. In this case, we have the following. Claim 2. If F1 is not complete, then F1 = (r − 1)P3 ∪ sK 1 for some integers r 1 and s 0. Proof of Claim 2. Since F1 is not complete, it follows that F1 contains two nonadjacent vertices x and y, which implies that |degG x − degG y| = 1. Assume, without loss of generality, that degG x = degG y + 1. Hence there exists a vertex z ∈ A1 such that yz ∈ / E(G) and xz ∈ E(G). Suppose that there is a vertex t ∈ A1 − {x, y, z}. It cannot occur that tx ∈ / E(G) and ty ∈ / E(G); for otherwise, {t, x, y} is independent in G. Thus t is adjacent to at least one of x and y. which implies that it is adjacent to both x and y. Consequently, t is adjacent to z as well. That is, every vertex of A1 different from x, y, and z is adjacent to all three of these vertices. This implies that F1 = (r − 1)P3 ∪ sK 1 for some integers r 1 and s 0, as claimed. Therefore, G = rP 3 ∪ sK 1 , where r 1 and s 0.
Fig. 3 shows a graph G of order 6 that satisfies the conditions described in Lemma 3.10. Observe that G = P3 ∪ 3K1 , where P3 : v, u, w and V (3K1 ) = {x, y, z}. Theorem 3.11. Let G be a connected graph of even order n = 2k 4 with diam G = 2 that is not a star. If (a) |N ∗ (u)| + |N ∗ (v)| = 0 for every two nonadjacent vertices u and v of G, or (b) |N ∗ (u)| + |N ∗ (v)| = 1 for every two nonadjacent vertices u and v of G, then G is not uniformly cordial. Proof. First, assume that G satisfies (a). Then for vertices u, v, and w, whenever uv, vw ∈ / E(G), we have uw ∈ / E(G); for otherwise, N ∗ (u, v) = ∅, contrary to hypothesis. This implies that G is a complete k-partite graph for some k 2. Since G is not a star, it follows by Propositions 2.1 and 3.3 that G is not uniformly cordial. Next, assume that G satisfies (b). By Lemma 3.10, G = rP 3 ∪ sK 1 , where r 1 and s 0, and so n = 3r + s = 2k for some integer k 2. We consider three cases. Case 1: 3r k. Partition V (G) into two subsets V0 and V1 with |V0 | = |V1 | = k such that V (rP 3 ) ⊆ V0 . Define a k − 2r = k(k − 1) − 2r and friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = 2 2 m1 (f ) = k 2 . Since c(f ) = k + 2r 4, it follows that f is not cordial. Case 2: 3r > k and k ≡ 0 (mod3). Then k = 3t for some t 1. Partition V (G) into two subsets V0 and V1 such that |V0 | = |V1 | = kand V (tP 3 ) ⊆ V0 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then k m0 (f ) = 2 − 2r = k(k − 1) − 2r and m1 (f ) = k 2 . Since c(f ) = k + 2r 4, it follows that f is not cordial. 2
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Case 3: 3r > k and k ≡ / 0 (mod 3). Let P3 : v1 , v2 , v3 in G and let r = k/3. Partition V (G) into two subsets V0 and V1 with |V0 | = |V1 | = k such that V0 consists of V (r P3 ) and v1 , or V0 consists of V (r P3 ) and {v1 , v2 }, according to whether k ≡ 1 (mod 3) ork ≡ 2 (mod 3). Define a friendly labeling f by assigning i to every vertex in Vi for k − 2(r − 1) − 1 and m1 (f ) = k 2 − 1. Since c(f ) = k + 2r − 2 2, it follows that f is i = 0, 1. Then m0 (f ) = 2 2 not cordial. In order to determine which of the remaining connected graphs of even order and diameter 2 are uniformly cordial, we first present two lemmas. Lemma 3.12. If G = (K p + Ks ) + e, where p 3, s 1, and p + s is even, then G is not uniformly cordial. Proof. Let X = V (K p ) = {x1 , x2 , . . . , xp }, Y = {y1 , y2 , . . . , ys }, p + s = 2k 4, and e = x1 x2 . We consider two cases. Case 1: s = 1. Partition V (G) into V0 and V1 with |V0 | = |V1 | = k such that x1 , y1 ⊆ V0 and x2 ∈ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = k − 1 and m1 (f ) = k + 1. Since c(f ) = 2, it follows that f is not cordial, a contradiction. Case 2: s 2. There are two subcases. Subcase 2.1: p k + 1. Partition V (G) into V0 and V1 with |V0 | = |V1 | = k such that x1, x2 ∈ V0 ⊆ X. Define a friendly labeling f by assigning i to every vertex in Vi for i =0, 1. Then m0 (f )=1+(k−s)s + 2s =(2+2sk−s 2 −s)/2 and m1 (f ) = sk. Since c(f ) = (s 2 + s − 2)/2 2, it follows that f is not cordial, a contradiction. Subcase 2.2: p k. Partition V (G) into V0 and V1 with |V0 | = |V1 | = k such ⊆ V0 . Define a friendly thatX k p labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = 1 + 2 − and m1 (f ) = k 2 . Since 2 2 p p c(f ) = k + − 1 1 + 4, it follows that f is not cordial, a contradiction. 2 2 Lemma 3.13. Let G be a connected graph of even order n 4 with diam G = 2. If {|N ∗ (u)| + |N ∗ (v)| : u, v ∈ V (G)
and
uv ∈ / E(G)} = {0, 1},
then G = (rK 1 ∪ K2 ) + F for some integer r 1 and for some graph F for which {|NF∗ (u)| + |NF∗ (v)| : u, v ∈ V (F ) and
uv ∈ / E(F )} ⊆ {0, 1}.
Proof. Let u and v be two nonadjacent vertices of G such that |N ∗ (u)|+|N ∗ (v)|=1, say N ∗ (u, v)=∅ and N ∗ (v, u)={w} for some vertex w ∈ V (G). Necessarily, w is not adjacent to u. However, w is adjacent to every vertex in N (u, v); for otherwise, |N ∗ (u)| + |N ∗ (v)|2. Let X = {x ∈ V (G) : ux ∈ / E(G) and |N ∗ (u, x)| + |N ∗ (x, u)| = 1}, Y = {y ∈ V (G) : uy ∈ / E(G) and |N ∗ (u, y)| + |N ∗ (y, u)| = 0}. Thus v ∈ X and each vertex in Y is adjacent to every vertex in N (u, v) = N (u). We first verify the following two claims. Claim 1. X = {v, w}. Proof of Claim 1. First we show that {v, w} ⊆ X. We already observed that v ∈ X. Since N ∗ (v, u) = {w}, it follows that vw ∈ E(G) and uw ∈ / E(G). Thus v ∈ N ∗ (w, u) and so |N ∗ (u, w)| + |N ∗ (w, u)| = 1. Thus w ∈ X. Hence {v, w} ⊆ X. Next, we show that X ⊆ {v, w}. Assume, to the contrary, that X{v, w}. Then there exists x ∈ X − {v, w}. So x is not adjacent to u and |N ∗ (u, x)| + |N ∗ (x, u)| = 1. First, we show that x is adjacent to neither v nor w. Assume, without loss of generality, that x is adjacent to v. Hence w, x ∈ N ∗ (v, u), which is impossible. We now complete the proof of Claim 1 by considering two cases.
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Case 1: N ∗ (u, x) = ∅. Then N ∗ (u, x) = {z} for some z ∈ V (G). Thus z is adjacent to u and z is not adjacent to x. Consider the two nonadjacent vertices x and z. Since (1) z is adjacent to v and w and (2) x is adjacent to neither v nor w, it follows that v, w ∈ N ∗ (z, x), which is impossible. Case 2: N ∗ (x, u) = ∅. Let N ∗ (x, u) = {z }. Thus z is adjacent to x and z is not adjacent to u. If z is adjacent to v, then w, z ∈ N ∗ (v, u), a contradiction. Thus z is not adjacent to v. Similarly, z is not adjacent to w. We now consider the two nonadjacent vertices v and x. Since w ∈ N ∗ (v, x) and z ∈ N ∗ (x, v), a contradiction is produced. Therefore, X = {v, w}, which completes the proof of Claim 1. Claim 2. Y ∪ {u} is an independent set in G. Proof of Claim 2. Certainly, Claim 2 is true if Y = ∅. Thus we may assume that Y = ∅. Assume, to the contrary, Y ∪ {u} is not independent in G. Since u is not adjacent to any vertex in Y , there exist adjacent vertices y, y ∈ Y . However, then, y ∈ N ∗ (y, u), contradicting that N ∗ (y, u) = ∅. This completes the proof of Claim 2. Since vw ∈ E(G), it follows by Claim 1 that X = K2 . By Claim 2, we have Y ∪ {u} = rK 1 for some integer r 1. Let F = N (u, v). Since each vertex in X ∪Y ∪{u} is adjacent to every vertex in F , it follows that G=(rK 1 ∪K2 )+F . If u ∈ V (F ), then NF (u) = NG (u) − V (rK 1 ∪ K2 ). Hence, if u and v are two nonadjacent vertices of F , then ∗ (u) and N ∗ (v) = N ∗ (v). This implies that NF∗ (u) = NG F G ∗ ∗ |NF∗ (u)| + |NF∗ (v)| = |NG (u)| + |NG (v)|1
for every two nonadjacent vertices u and v in F .
Theorem 3.14. Let G be a connected graph of even order n 4 with diam G = 2. If {|N ∗ (u)| + |N ∗ (v)| : u, v ∈ V (G)
and uv ∈ / E(G)} = {0, 1},
(1)
then G is not uniformly cordial. Proof. Assume, to the contrary, that there exist connected uniformly cordial graphs of even order at least 4 with diameter 2 that satisfy (1). Among these graphs, let G be one of minimum order n=2k 4. By Lemma 3.13, G=(rK 1 ∪K2 )+F for some integer r 1 and some graph F for which {|NF∗ (u)| + |NF∗ (v)| : u, v ∈ V (F )
and
uv ∈ / E(F )} ⊆ {0, 1}.
(2)
We consider two cases. Case 1: F is disconnected. There are two subcases. Subcase 1.1: F is empty. Then F = tK 1 for some integer t 2 and so G = (rK 1 ∪ K2 ) + tK 1 . Let s=r+2 3, and let Ks,t be the complete bipartite graph with partite sets X={x1 , x2 , . . . , xs } and Y ={y1 , y2 , . . . , yt }. Then G is obtained from Ks,t by adding an edge joining two vertices of X, say x1 x2 . There are two subcases. Subcase 1.1.1: 3s t. Partition V (G) into V0 and V1 such that |V0 | = |V1 | = k and X ⊆ V0 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = s(k − s) + 1 = sk − s 2 + 1 and m1 (f ) = sk. Since c(f ) = s 2 − 18, it follows that f is not cordial, a contradiction. Subcase 1.1.2: 2 t < s. Partition V (G) into V0 and V1 with |V0 | = |V1 | = k such that Y ∪ {x1 } ⊆ V0 and x2 ∈ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = t (k − t) = tk − t 2 and m1 (f ) = tk + 1. Since c(f ) = 1 + t 2 5, it follows that f is not cordial, a contradiction. Subcase 1.2: F is not empty. We claim that F =tK 1 ∪K2 for some integer t 1; otherwise, there exist two nonadjacent vertices u and v in F with |NF∗ (u)| + |NF∗ (v)|2, contradiction property (2) possessed by F . Hence, as claimed, G = (rK 1 ∪ K2 ) + (tK 1 ∪ K2 ). Let p = r + 2 3, q = t + 2 3, and let Kp,q be the complete bipartite graph with partite sets X = {x1 , x2 , . . . , xp } and Y = {y1 , y2 , . . . , yq }. We may assume that p q and that G is obtained from Ks,t by adding the two edges x1 x2 and y1 y2 . There are two subcases.
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Fig. 4. The graph G = (rK 1 ∪ K2 ) + F .
Subcase 1.2.1: p = q. Define a friendly labeling f by assigning 0 to every vertex in X and 1 to every vertex in Y . Then m0 (f ) = 2 and m1 (f ) = k 2 . Since c(f ) = k 2 − 2 7, it follows that f is not cordial, a contradiction. Subcase 1.2.2: p < q. Partition V (G) into two subsets V0 and V1 with |V0 |=|V1 |=k such that X∪{y1 } ⊆ V0 and y2 ∈ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i =0, 1. Then m0 (f )=p(k −p)+1=pk −p2 +1 and m1 (f ) = pk + 1. Since c(f ) = p 2 9, it follows that f is not cordial, a contradiction. Case 2: F is connected. There are two subcases. Subcase 2.1: F is complete. Then G = (rK 1 ∪ K2 ) + Ks = (K r+2 + Ks ) + e for some integer s 1. It follows by Lemma 3.12 that G is not uniformly cordial, a contradiction. Subcase 2.2: F is not complete. Then diam F = 2; for otherwise, there exist two vertices z1 and z2 in F such that dF (z1 , z2 )3, implying that |NF∗ (z1 )| + NF∗ (z2 )|2, contradicting property (2) of F . Let V (K2 ) = {x, y}, where K2 is the subgraph in G = (rK 1 ∪ K2 ) + F (see Fig. 4). We consider two cases. Subcase 2.1.1: F has odd order. Let V (F ) = 2p + 13 and |V (G) − V (F )| = 2q + 1 3. Since F = K3 , it follows that F is not uniformly cordial. So there is a friendly labeling f of F that is not cordial. We may assume that Ui is the set of vertices of F labeled i by f for i = 0, 1, where |U0 | = p + 1 and |U1 | = p. Partition V (G) − V (F ) into two subsets V0 and V1 with |V0 | = q and |V1 | = q + 1 such that {x, y} ⊆ V1 . Define a friendly labeling f of G by f (v) = f (v) if v ∈ V (F ) and f (v) = i if v ∈ Ui for i = 0, 1. Then m0 (f ) = m0 (f ) + q(p + 1) + p(q + 1) + 1 and m1 (f ) = m1 (f ) + qp + (q + 1)(p + 1). Since c(f ) = c(f ) 2, it follows that f is not cordial. Subcase 2.1.2: F has even order. Let |V (F )| = 2p 4 and |V (G) − V (F )| = 2q = 2k − 2p 4. Since G is a uniformly cordial graph of minimum even order n 4 with diameter 2 that satisfies (1) and the order of F is smaller than that of G, it follows that F is not uniformly cordial. Thus there is a friendly labeling f of F that is not cordial. Assume first that m0 (f ) > m1 (f ). So c(f ) = m0 (f ) − m1 (f )2. Let Ui be the set of vertices of F labeled i by f for i = 0, 1, where |U0 | = |U1 | = p. Partition V (G) − V (F ) into two subsets V0 and V1 with |V0 | = |V1 | = q such that {x, y} ⊆ V1 . Define a friendly labeling f of G by f (v) = f (v) if v ∈ V (F ) and f (v) = i if v ∈ Ui for i = 0, 1. Then m0 (f ) = m0 (f ) + pq + pq + 1 and m1 (f ) = m1 (f ) + pq + pq. Since c(f ) = m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) + 1 = c(f ) + 1 3, it follows that f is not cordial. Next, assume that m1 (f ) − m0 (f )2. Let Ui be the set of vertices of F labeled i by f for i = 0, 1, where |U0 | = |U1 | = p. Partition V (G) − V (F ) into two subsets V0 and V1 with |V0 | = |V1 | = q such that x ∈ V0 and y ∈ V1 . Define a friendly labeling f of G by f (v) = f (v) if v ∈ V (F ) and f (v) = i if v ∈ Ui for i = 0, 1. Then m0 (f ) = m0 (f ) + pq + pq and m1 (f ) = m1 (f ) + pq + pq + 1. Since c(f ) = m1 (f ) − m0 (f ) = m1 (f ) − m0 (f ) + 1 = c(f ) + 1 3, it follows that f is not cordial.
By Theorems 3.11 and 3.14, we have the following.
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Corollary 3.15. Let G be a connected graph of even order n 4 with diam G = 2 that is not a star. If |N ∗ (u)| + |N ∗ (v)| 1 for every pair u, v of nonadjacent vertices of G, then G is not uniformly cordial. By Proposition 2.1 and Corollaries 3.9 and 3.15, we have the following. Corollary 3.16. Let G be a connected graph of even order. Then G is uniformly cordial if and only if G is a star. Combining Corollaries 3.2 and 3.16, we now present a characterization of uniformly cordial graphs. Theorem 3.17. A nontrivial connected graph G of order n is uniformly cordial if and only if (1) n = 3 and G = K3 , or (2) n is even and G = K1,n−1 . Acknowledgments We are grateful to the referee whose valuable suggestions resulted in an improved paper. References [1] M. Benson, S.M. Lee, On cordialness of regular windmill graphs, Congr. Numer. 68 (1989) 49–58. [2] I. Cahit, Cordial graphs: A weaker version of graceful and harmonious graphs, Ars Combin. 23 (1987) 201–207. [3] I. Cahit, Recent results and open problems on cordial graphs, in: Contemporary Methods in Graph Theory, Bibligraphisches Inst., Mannhiem, 1990, pp. 209–230. [4] G. Chartrand, P. Zhang, Introduction to Graph Theory, McGraw-Hill, Boston, 2005. [5] G.M. Du, Cordiality of complete k-partite graphs and some special graphs, Neimenggu Shida Xuebao Ziran Kexue Hanwen Ban 2 (1997) 9–12. [6] J.A. Gallian, A dynamic survey of graph labeling, Electron. J. Combin. 5 (2000) 1–79. [7] Y.S. Ho, S.M. Lee, S.S. Shee, Cordial labelings of unicyclic graphs and generalized Petersen graphs, Congr. Numer. 68 (1989) 109–122. [8] Y.S. Ho, S.M. Lee, S.S. Shee, Cordial labelings of the Cartesian product and composition of graphs, Ars Combin. 29 (1990) 169–180. [9] W.W. Kirchherr, On the cordiality of certain specific graphs, Ars Combin. 31 (1991) 127–138. [10] W. W. Kirchherr, Algebraic approaches to cordial labeling, Graph Theory, Combinatorics, Algorithms, and Applications, Y. Alavi, et al. (Eds.) SIAM, Philadelphia, PA, 1991, pp. 294–299. [11] W.W. Kirchherr, NEPS operations on cordial graphs, Discrete Math. 115 (1993) 201–209. [12] S. Kuo, G.J. Chang, Y.H.H. Kwong, Cordial labeling of mK n , Discrete Math. 169 (1997) 1–3. [13] S.M. Lee, A. Liu, A construction of cordial graphs from smaller cordial graphs, Ars Combin. 32 (1991) 209–214. [14] Y.H. Lee, H.M. Lee, G.J. Chang, Cordial labelings of graphs, Chinese J. Math. 20 (1992) 263–273; Y.H. Lee, H.M. Lee, G.J. Chang, Cordial labelings of graphs, Chinese J. Math. 20 (1962) 270–288 [15] E. Seah, On the construction of cordial graphs, Ars. Combin. 31 (1991) 249–254. [16] S.C. Shee, The cordiality of the path-union of n copies of a graph, Discrete Math. 151 (1996) 221–229. [17] S.C. Shee, Y.S. Ho, The cordiality of the one-point union of n-copies of a graph, Discrete Math. 117 (1993) 225–243.
Uniformly cordial graphs Gary Chartranda , Sin-Min Leeb , Ping Zhanga,1 a Department of Mathematics, Western Michigan University, Kalamazoo, MI 49008, USA b Department of Computer Sciences, San Jose State University, San Jose, CA 95192, USA
Received 24 May 2002; received in revised form 7 November 2004; accepted 7 November 2005 Available online 31 March 2006 To the Memory of W. T. Tutte
Abstract Let G be a graph with vertex set V (G) and edge set E(G). A labeling f : V (G) → {0, 1} induces an edge labeling f ∗ : E(G) → {0, 1}, defined by f ∗ (xy) = |f (x) − f (y)| for each edge xy ∈ E(G). For i ∈ {0, 1}, let ni (f ) = |{v ∈ V (G) : f (v) = i}| and mi (f ) = |{e ∈ E(G) : f ∗ (e) = i}|. Let c(f ) = |m0 (f ) − m1 (f )|. A labeling f of a graph G is called friendly if |n0 (f ) − n1 (f )| 1. A cordial labeling of G is a friendly labeling f for which c(f ) 1. A graph G is a uniformly cordial graph if every friendly labeling of G is cordial. It is shown that a connected graph G of order n 2 is uniformly cordial if and only if n = 3 and G = K3 , or n is even and G = K1,n−1 . © 2006 Elsevier B.V. All rights reserved. Keywords: Friendly labeling; Cordial labeling; Uniformly cordial graph
1. Introduction We refer to the book [4] for graph theory notation and terminology not described in this paper. Let G be a graph with vertex set V (G) and edge set E(G). A labeling f : V (G) → {0, 1} induces an edge labeling f ∗ : E(G) → {0, 1}, defined by f ∗ (xy) = |f (x) − f (y)| for each edge xy ∈ E(G). For i ∈ {0, 1}, let ni (f ) = |{v ∈ V (G) : f (v) = i}| and
mi (f ) = |{e ∈ E(G) : f ∗ (e) = i}|.
Let c(f ) = |m0 (f ) − m1 (f )|. A labeling f of a graph G is called friendly if |n1 (f ) − n0 (f )| 1. A cordial labeling is a friendly labeling f for which c(f )1. Note that interchanging the vertex labels 0 and 1 in a cordial labeling results in a new cordial labeling of G. A cordial graph is a graph that admits a cordial labeling. Cordial labelings of graphs were introduced by Cahit [2], who showed that (1) every tree is cordial, (2) Kn is cordial if and only if n 3, (3) Kr,s is cordial for all r and s, (4) the wheel Wn is cordial if and only if n ≡ 3 (mod 4), (5) Cn is cordial if and only if n = 2 (mod 4), and (6) an Eulerian graph is not cordial if its size is congruent to 2 modulo 4. Benson and Lee [1] described a large class of regular cordial windmill graphs. Du [5] investigated cordial complete k-partite graphs. Kuo et al. [12] determined all m and n for which mK n is cordial. Lee, et al. [14] exhibited some cordial graphs. Generalized Petersen graphs that are cordial are characterized in [7]. Ho et al. [8] investigated the construction 1 Research supported in part by a Western Michigan University Faculty Research and Creative Activities Fund.
E-mail address: [email protected] (P. Zhang). 0012-365X/$ - see front matter © 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2005.11.025
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of cordial graphs using Cartesian products and compositions of graphs. Shee and Ho [17] determined the cordiality of (n) Cm , the one-point union of n copies of Cm . Several constructions of cordial graphs were proposed in [13,15–17,9–11]. Other results and open problems concerning cordial graphs are given in [3,6]. A graph G is defined to be uniformly cordial if every friendly labeling of G is cordial. That is, a graph G is uniformly cordial if whenever V (G) is partitioned at random into two subsets whose cardinalities are as equal as possible, then this induces a cordial labeling of G. For example, K3 is uniformly cordial. Since Kn , n4, is not cordial, it is not uniformly cordial either. Proposition 1.1. The complete graph Kn , n2, is a uniformly cordial graph if and only if n 3. 2. Some preliminary results We first determine all uniformly cordial complete bipartite graphs. Proposition 2.1. A complete bipartite graph Kr,s , where 1 r s, is uniformly cordial if and only if r = 1 and s is odd. Proof. Let G = Kr,s , where 1 r s. First, assume that r = 1 and s is odd. Then s = 2k − 1 for some integer k 1. Let f be a friendly labeling of G, where Vi is the set of vertices of G labeled i for i = 0, 1. Then |V0 | = |V1 | = k. Let u be the central vertex of G. Assume, without loss of generality, that u ∈ V0 . Then m0 (f ) = k − 1 and m1 (f ) = k. Since c(f ) = 1, it follows that f is cordial. Therefore, G is uniformly cordial. For the converse, assume that G = K1,s for any odd integer s. We consider two cases. Case 1: G is a star. Then r = 1 and s is even. Let s = 2k for some integer k 1. Let V (G) = {u, u1 , u2 , . . . , us }, where u is the central vertex of G. Define a labeling f of G by f (u) = f (ui ) = 0 for 1i k − 1, and f (ui ) = 1 for k i 2k. Since n0 (f ) = k and n1 (f ) = k + 1, it follows that f is a friendly labeling of G. On the other hand, m0 (f ) = k − 1 and m1 = k + 1. Thus c(f ) = 2 and so f is not cordial. Therefore, G is not uniformly cordial. Case 2: G is not a star. Then 2 r s. Let X and Y be the partite sets of G with |X| = r and |Y | = s. If r = s, then we define a friendly labeling f of G by assigning 0 to each vertex in X and 1 to each vertex in Y . Then m0 (f ) = 0 and m1 (f ) = r 2 . Since c(f ) = r 2 4, it follows that f is not cordial. Thus we may assume that 2 r < s. Partition V (G) into V0 and V1 such that V0 | − |V1 1 and X ⊆ V0 . Define a friendly labeling f of G by assigning i to every vertex in Vi for i = 0, 1. There are two cases. Subcase 2.1: r + s is even. Then r + s = 2k for some integer k 2. Thus |V0 | = |V1 | = k. Then m0 (f ) = r(k − r) and m1 (f ) = rk. Since c(f ) = r 2 4, it follows that f is not cordial. Subcase 2.2: r + s is odd. Then r + s = 2k + 1 for some integer k 2. Let |V0 | = k and |V1 | = k + 1. Then m0 (f ) = r(k − r) and m1 (f ) = r(k + 1). Since c(f ) = r 2 + r 6, it follows that f is not cordial. Therefore, G is not uniformly cordial in this case. It will be useful to know that certain classes of graphs are not uniformly cordial. Lemma 2.2. Let n3. If G is obtained from Kn−1 by adding a pendant edge, then G is not uniformly cordial. Proof. Suppose that G is obtained from Kn−1 by adding the pendant edge uv, where u ∈ V (Kn−1 ). We consider two cases. Case 1: n is odd. Then n = 2k + 1 for some integer k 1. Partition V (Kn−1 ) into V0 and V1 such that |V0 | = |V1 | = k and u ∈ V0 . Define a friendly labeling f of G by assigning i to every vertex in Vi for i = 0, 1 and assigning 1 to the k = k 2 − k and m1 (f ) = k 2 + 1. Since |m0 (f ) − m1 (f )| = k + 1 2, it follows that f vertex v. Then m0 (f ) = 2 2 is not cordial. Case 2: n is even. The graph G is not uniformly cordial for n = 4 and n = 6. Friendly labelings of G that are not cordial are shown in Fig. 1 for n = 4 and n = 6. For n 8, let n=2k, where then k 4. Now partition V (G) into V0 and V1 such that |V0 |=|V1 |=kandu, v ∈ V0 . Define k + +1=k 2 −2k +2 a friendly labeling f of G by assigning i to every vertex in Vi for i =0, 1. Then m0 (f )= k−1 2 2
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Fig. 1. Friendly labelings that are not cordial for n = 4, 6.
and m1 (f ) = (k − 1)k = k 2 − k. Since k 4, it follows that |m0 (f ) − m1 (f )| = k − 2 2 and so f is not cordial. Lemma 2.3. Let n 5. If G = Kn − M for some matching M in Kn , then G is not uniformly cordial. Proof. Let M = {e1 , e2 , . . . , e }, where ei = xi yi for 1 i . We consider two cases. Case 1: n is odd. Let n = 2k + 1, where then k 2. Partition V (G) into V0 and V1 such that (1) V0 | − |V1 = 1 and (2) xi and yi both belong to V0 or xi and yi both belong to V1 for eachi (1i ). Define a friendly labeling f of G by k + − = k 2 − and m1 (f ) = k(k + 1) = k 2 + k. assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = k+1 2 2 Since c(f ) = k + 3, it follows that f is not cordial. Case 2: n is even. Let n = 2k, where then k 3. There are two subcases. Subcase 2.1: k is odd and =k. Let k=2a+1. Partition V (G) into V0 and V1 such that V0 ={xi , yi : 1 i a}∪{x2a+1 } and V1 = {xi , yi : a + 1i 2a} ∪ {y2a+1 }. Then |V0 | = |V1 | = k. Define a friendly labeling f of G by assigning i to k every vertex in Vi for i =0, 1. Then m0 (f )=2 −(k −1)=(k −1)2 and m1 (f )=k 2 −1. Since c(f )=2(k −1) 4, 2 it follows that f is not cordial. Subcase 2.2: Subcase 2.1. does not occur. Partition V (G) into V0 and V1 such that (1) |V0 | = |V1 | = k and (2) xi ∗ and yi both belong to V0 or xi and yi both belong to V1 for each i (1 i). Define a friendly labeling f of G by k k assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ∗ ) = + − = k(k − 1) − and m1 (f ∗ ) = k 2 . Since 2 2 c(f ∗ ) = k + 4, it follows that f ∗ is not cordial. 3. Which graphs are randomly cordial? In an attempt to characterize all connected graphs that are uniformly cordial, we begin by showing that all connected graphs (of sufficiently large odd order) are not uniformly cordial. For a vertex v in a graph G, let N (v) denote the neighborhood of v and N [v] = N (v) ∪ {v}. 3.1. Graphs with odd order Theorem 3.1. No connected graph of odd order n 5 is a uniformly cordial graph. Proof. Assume, to the contrary, that there is a connected graph G of order n = 2k + 1 5 that is uniformly cordial. By Propositions 1.1 and 2.1 and Lemmas 2.2 and 2.3, we may assume that G is not the complete graph Kn , the star K1,n−1 , the graph Kn−1 with a pendant edge, or Kn − M for some matching M in G. Thus, there exists a vertex u in G such that 2 deg u = n − 3 = 2k − 2. Let N (u) = {u1 , u2 , . . . , u } and V (G) − N [u] = {v1 , v2 , . . . , v2k− }. Consider the sequence s : u, u1 , u2 , . . . , u , v1 , v2 , . . . , v2k− of the vertices of G. Partition V (G) into two subsets V0
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and V1 such that V0 consists of the first k + 1 vertices of s. Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. By assumption, f is a cordial labeling of G and so c(f ) = |m0 (f ) − m1 (f )| 1. Define a labeling f of G by f (u) = 1 and f (v) = f (v) for all v ∈ V (G) − {u}. Then n0 (f ) = k and n1 (f ) = k + 1, implying that f is a friendly labeling of G as well. We show that f is not a cordial labeling of G. There are two cases. Case 1: k. Then m0 (f ) = m0 (f ) − and m1 (f ) = m1 (f ) + . Thus m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) − 2. Since c(f )1 and 24, it follows that c(f ) 3. Case 2: k+1. Then m0 (f )=m0 (f )−k+(−k)=m0 (f )+−2k and m1 (f )=m1 (f )+k−(−k)=m1 (f )−+2k. Thus m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) + 2 − 4k = m0 (f ) − m1 (f ) − 2(2k − ). Since c(f )1 and 2k − 2, it follows that c(f ) 3. In either case, f is not a cordial labeling of G, which is a contradiction.
Corollary 3.2. Let G be a connected graph of odd order n3. Then G is a uniformly cordial graph if and only if G = K3 . In Proposition 2.1, we determined all uniformly cordial bipartite graphs. Theorem 3.1 can be used to show that, with the exception of K3 , there are no complete multipartite graphs that are not stars. Proposition 3.3. For k 3, no complete k-partite graph different from K3 is uniformly cordial. Proof. Assume, to the contrary, that there is a complete k-partite G with G = K3 that is uniformly cordial, where k 3. Let G = Kn1 ,n2 ,...,nk , where n1 n2 · · · nk and n = n1 + n2 + · · · + nk . By Theorem 3.1, n is even. Let n = 2t for some integer t 2 and let X1 , X2 , . . . , Xk be the partite sets of G such that |Xi | = ni for 1 i k. We consider two cases. Case 1: n1 t + 1. Partition V (G) into V0 and V1 such that |V0 | = |V1 | = t and V0 ⊆ X1 . Then V1 = V11 ∪ V12 such that V11 = X1 ∩ V1 and |V12 | = 2. Define a friendly labelingf by assigning i to every vertex in Vi for i = 0, 1. Note that m0 (f ) = (t − ) and m1 (f ) = t. Since c(f ) = 2 4, it follows that f is not cordial, which is a contradiction. Case 2: n1 t. There are two subcases. Subcase 2.1: n2 2. Partition V (G) into V0 and V1 such that (1) |V0 |=|V1 |=t and (2) X1 ⊂ V0 and X2 ⊆ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. By assumption, f is a cordial labeling of G and so c(f ) = |m0 (f ) − m1 (f )|1. Let u ∈ X1 and v ∈ X2 . Thus f (u) = 0 and f (v) = 1. Define a new friendly labeling f by f (u) = 1, f (v) = 0, and f (x) = f (x) for each x ∈ V (G) − {u, v}. Now m0 (f ) = m0 (f ) + (n1 − 1) + (n2 − 1). Therefore, m1 (f )=mi (f )−(n1 −1)−(n2 −1). It follows then that m1 (f )−m0 (f )=m1 (f )−m0 (f )−2(n1 +n2 −2). Since n1 + n2 − 2 2 and c(f )1, it follows that c(f ) 3 and so f is not cordial, a contradiction. Subcase 2.2: n2 = 1. Hence n1 t and ni = 1 for 2 i k. Since G is not a complete graph, n1 2. This implies that G = K n1 + Kn−n1 . Partition V (G) into V0 and V1 such that |V0 | = |V1 | = t and X 1 ⊆ V0 . Define a friendly labeling f ∗ of G by assigning i to each vertex in Vi for i = 0, 1. Then m0 (f ∗ ) = 2 2t − n21 and m1 (f ∗ ) = t 2 . Thus c(f ∗ ) = t + n21 3, implying that f ∗ is not cordial. Therefore, in either case, G is not uniformly cordial. 3.2. Graphs of even order with diameter at least 3 We now know that if there is any uniformly cordial graph G that is neither K3 nor a star, then G must have even order. We have seen in Proposition 1.1 that K3 is the only complete graph that is uniformly cordial. Since complete graphs are the only connected graphs with diameter 1, we assume that all connected graphs under consideration have even order and diameter at least 2. In this section, we show that there exists no uniformly cordial graph of even order with diameter 3 or more. In order to do this, we first show that there are certain properties that no uniformly cordial graphs of even order may possess.
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For two nonadjacent vertices u and v in a graph G, let N (u, v) = N (u) ∩ N (v), N ∗ (u, v) = N (u) − N (u, v), and If the two nonadjacent vertices u and v under discussion are clear, then we simply write N ∗ (u, v) = N ∗ (u) and N ∗ (v, u) = N ∗ (v). N ∗ (v, u) = N(v) − N (u, v).
Lemma 3.4. Let G be a connected graph of even order 2k 4. If G contains two nonadjacent vertices u and v such that (1) |N ∗ (u)| + |N ∗ (v)|2 and (2) |N ∗ (u)|k − 1 and |N ∗ (v)| k − 1, then G is not uniformly cordial. Proof. Let |N ∗ (u)| = a and |N ∗ (v)| = b. Partition V (G) into two subsets V0 and V1 with |V0 | = |V1 | = k such that {u} ∪ N ∗ (u) ⊆ V0 and {v} ∪ N ∗ (v) ⊆ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. By assumption, f is a cordial labeling of G and so c(f ) = |m0 (f ) − m1 (f )| 1. Define a new friendly labeling f by f (u) = 1, f (v) = 0, and f (x) = f (x) for all x ∈ V (G) − {u, v}. Since m0 (f ) = m0 (f ) − a − b
and
m1 (f ) = m1 (f ) + a + b,
it follows that m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) − 2(a + b). Since a + b 2 and c(f )1, it follows that c(f ) 3. Therefore, G is not uniformly cordial.
Lemma 3.5. Let G be a connected graph of even order 2k 4. If G contains two vertices u and v such that |N ∗ (u)| k and |N ∗ (u)| − |N ∗ (v)|2k − 4, then G is not uniformly cordial. Proof. Let |N ∗ (u)| = a and |N ∗ (v)| = b. Thus a k and a − b 2k − 4. Partition V (G) into two subsets V0 and V1 with |V0 | = |V1 | = k such that V0 consists of u and k − 1 vertices from N ∗ (u) − {v}. Then v ∈ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. By assumption, f is a cordial labeling of G and so c(f ) = |m0 (f ) − m1 (f )|1. Define a new friendly labeling f by f (u) = 1, f (v) = 0, and f (x) = f (x) for all x ∈ V (G) − {u, v}. If u and v are nonadjacent, then m0 (f ) = m0 (f ) − (k − 1) + [a − (k − 1)] − b, m1 (f ) = m1 (f ) + (k − 1) − [a − (k − 1)] + b. If u and v are adjacent, then m0 (f ) = m0 (f ) − (k − 1) + [(a − 1) − (k − 1)] − (b − 1), m1 (f ) = m1 (f ) + (k − 1) − [(a − 1) − (k − 1)] + (b − 1). In either case, m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) − 2[2(k − 1) − a + b]. Note that a − b 2k − 4 and so 2(k − 1) − a + b 2. Since c(f ) 1, it follows that c(f ) 3. Therefore, G is not uniformly cordial. Proposition 3.6. Let G be a connected graph of even order 2k 4. If G contains two nonadjacent vertices u and v such that (1) N (u, v) = ∅ or (2) N ∗ (u) = ∅ and N ∗ (v) = ∅, then G is not uniformly cordial. Proof. Let |N ∗ (u)| = a and |N ∗ (v)| = b. We consider two cases. Case 1: N (u, v) = ∅. Then N ∗ (u) = N (u) and N ∗ (v) = N (v). Since (1) G is connected, (2) u and v are nonadjacent, and (3) N (u, v) = ∅, it follows that 1 a 2k − 3
and
1b 2k − 3.
If a k − 1 and b k − 1, then G is not uniformly cordial by Lemma 3.4 since a + b 2. Thus, assume that one of a and b is at least k, say a k. Since a − b (2k − 3) − 1 = 2k − 4, it follows by Lemma 3.5 that G is not uniformly cordial.
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Case 2: N ∗ (u) = ∅ and N ∗ (v) = ∅. Thus, a, b 1. Since u and v are nonadjacent, it follows that 1 a 2k − 3 and 1b 2k − 3. An argument similar to one used in Case (1) shows that G is not uniformly cordial. The distance d(u, v) between two vertices u and v in a connected graph G is the length of a shortest u − v path in G. For a vertex u in a connected graph G, the eccentricity e(u) of u is the distance between u and a vertex of G that is farthest from u. The minimum eccentricity among the vertices of G is its radius rad G and the maximum eccentricity is its diameter diam G. Corollary 3.7. Let G be a connected graph of even order n4. If diam G3, then G is not uniformly cordial. Proof. Let u and v be two vertices of G such that d(u, v) 3. Thus u and v are nonadjacent and N (u, v) = ∅. It then follows by Proposition 3.6 that G is not uniformly cordial. 3.3. Graphs of even order with diameter 2 By Corollary 3.7 no connected graph of even order with diameter 3 or more is uniformly cordial. We now consider connected graphs of even order with diameter 2. We first show that there are certain properties that no uniformly cordial graph of even order having diameter 2 may possess. Some additional definitions and notation will be useful. For a graph G and a set S of vertices of G, let G be the complement of G and let S be the subgraph of G induced by S. For two vertex-disjoint graphs G and H , let G ∪ H be the union of G and H and let G + H be the join of G and H . Lemma 3.8. Let G be a connected graph of even order n = 2k 4 with diam G = 2. If G contains two nonadjacent vertices u and v such that |N ∗ (u)|k and N ∗ (v) = ∅, then G is not uniformly cordial. Proof. If |N ∗ (u)| 2k − 4, then G is not uniformly cordial by Lemma 3.5. Thus we may assume that |N ∗ (u)| 2k − 3. Since diam G = 2, it follows that N (u, v) = ∅, which implies that |N ∗ (u)| 2k − 3. Therefore, |N ∗ (u)| = 2k − 3 and |N(u, v)| = 1. Let N (u, v) = {w}. Again, since diam G = 2, the vertex w is adjacent to every vertex in N ∗ (u). Let H = N ∗ (u). We consider three cases. Case 1: H is complete. Then G is obtained from Kn−1 by adding a pendant edge vw. Since n4, it then follows by Lemma 2.2 that G is not uniformly cordial. Case 2: H is an empty graph. Then G is obtained from K1,1,2k−3 , where u and w are two vertices of eccentricity 1 in K1,1,2k−3 , by adding a new vertex v and the edge vw. If k = 2, then G is obtained by adding a pendant edge to K3 and so G is not uniformly cordial by Lemma 2.2. For k 3, partition V (G) into V0 and V1 with |V0 | = |V1 | = k such that u, v, w ∈ V0 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = 2 + 2(k − 3) = 2k − 4
and
m1 (f ) = 2k.
Since c(f ) = 4, it follows that G is not uniformly cordial. Case 3: H is neither complete nor empty. Then H contains three vertices x, y, and z such that xy ∈ / E(H ) and xz ∈ E(H ). Consider the two nonadjacent vertices v and x in G. Since z, u ∈ N ∗ (x, v) and v, w, x, y ∈ / N ∗ (x, v), it follows that 2 |N ∗ (x, v)| 2k − 4. There are two subcases. Subcase 3.1: |N ∗ (x, v)|k. Since |N ∗ (x, v)| 2k − 4, it follows by Lemma 3.5 that G is not uniformly cordial. Subcase 3.2: |N ∗ (x, v)|k − 1. Since |N ∗ (v, x)| = 0 k − 1 and |N ∗ (x, v)| + |N ∗ (v, x)| = |N ∗ (x, v)| 2, it follows by Lemma 3.4 that G is not uniformly cordial. We now have the following corollary.
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Corollary 3.9. Let G be a connected graph of even order n 4 with diam G=2. If G contains two nonadjacent vertices u and v with |N ∗ (u)| + |N ∗ (v)|2, then G is not uniformly cordial. Proof. Let n = 2k 4. We consider two cases. Case 1: |N ∗ (u)| k − 1 and |N ∗ (v)|k − 1. By Lemma 3.4, G is not uniformly cordial. Case 2: Exactly one of |N ∗ (u)| and |N ∗ (v)| is at least k, say |N ∗ (u)| k. There are two subcases. Subcase 2.1: |N ∗ (v)| = 0. By Lemma 3.8, G is not uniformly cordial. Subcase 2.2: |N ∗ (v)| 1. Since diam G = 2, it follows that N (u, v) = ∅ and so |N ∗ (u)| 2k − 4. By Lemma 3.5, G is not uniformly cordial. It remains to determine which connected graphs G of even order n 4 with diam G=2 such that |N ∗ (u)|+|N ∗ (v)| 1 for every two nonadjacent vertices u and v of G are uniformly cordial. We begin with the case where |N ∗ (u)| + |N ∗ (v)| is constant for every two nonadjacent vertices u and v of G. Lemma 3.10. Let G be a connected graph of even order n 4 with diam G = 2. If |N ∗ (u)| + |N ∗ (v)| = 1 for every two nonadjacent vertices u and v of G, then G = rP 3 ∪ sK 1 for some integers r 1 and s 0. Proof. Observe that if v1 and v2 are two nonadjacent vertices of G, then | deg v1 − deg v2 | = 1. Let u be a vertex of G with e(u) = 2. For i = 1, 2, let Ai be the set of vertices of G at distance i from u. Thus Ai = ∅ for i = 1, 2 and V (G) = {u} ∪ A1 ∪ A2 . Let v ∈ A2 . So uv ∈ / E(G). By hypotheses, |N ∗ (u)| + |N ∗ (v)| = 1 and so exactly one of N ∗ (u) ∗ and N (v) is empty. Without loss of generality, we assume that N ∗ (u) = ∅; otherwise, we interchange the roles of u and v. Let Fi = Ai for i = 1, 2. Claim 1. F2 = K2 . Proof of Claim 1. Since |N ∗ (u)| + |N ∗ (v)| = 1 and N ∗ (u) = ∅, there exists exactly one vertex, say w, in F2 that is adjacent to v in F2 . All vertices of G that are adjacent to u are necessarily adjacent to v as well, that is, u and v are mutually adjacent to the vertices in A1 . Likewise, u and w are not adjacent. Since v is adjacent to w but not to u, every vertex in A1 is adjacent to w as well (see Fig. 2). Since vw ∈ E(G) and v, w ∈ A2 , it follows that F2 contains K2 . Assume, to the contrary, F2 = K2 . Then there exists a vertex x ∈ A2 − {v, w} and vx ∈ / E(G). If x is adjacent to a vertex y ∈ A2 , then N (x) = {y} ∪ A1 and / E(G). So x is an isolated vertex in F2 . Since ux ∈ / E(G), it degG v = degG x; however, this is impossible since vx ∈ follows that |N ∗ (u)| + |N ∗ (x)| = 1. This implies that x is adjacent to all but one vertex y in A1 . Since (1) x is not
Fig. 2. A step in the proof of Claim 1.
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Fig. 3. A graph G = P3 ∪ 3K1 in 3.10 for r = 1 and s = 3.
adjacent to v and w and (2) y is adjacent to v and w, it follows that |N ∗ (x)| + |N ∗ (y)| 2, which is a contradiction. Therefore, Fi = A2 = K2 , as claimed. By Claim 1, we have G = (K1 ∪ K2 ) + F1 . If F1 is complete, then G = P3 ∪ (n − 3)K1 and we have the desired result. Thus we may assume that F1 is not complete. In this case, we have the following. Claim 2. If F1 is not complete, then F1 = (r − 1)P3 ∪ sK 1 for some integers r 1 and s 0. Proof of Claim 2. Since F1 is not complete, it follows that F1 contains two nonadjacent vertices x and y, which implies that |degG x − degG y| = 1. Assume, without loss of generality, that degG x = degG y + 1. Hence there exists a vertex z ∈ A1 such that yz ∈ / E(G) and xz ∈ E(G). Suppose that there is a vertex t ∈ A1 − {x, y, z}. It cannot occur that tx ∈ / E(G) and ty ∈ / E(G); for otherwise, {t, x, y} is independent in G. Thus t is adjacent to at least one of x and y. which implies that it is adjacent to both x and y. Consequently, t is adjacent to z as well. That is, every vertex of A1 different from x, y, and z is adjacent to all three of these vertices. This implies that F1 = (r − 1)P3 ∪ sK 1 for some integers r 1 and s 0, as claimed. Therefore, G = rP 3 ∪ sK 1 , where r 1 and s 0.
Fig. 3 shows a graph G of order 6 that satisfies the conditions described in Lemma 3.10. Observe that G = P3 ∪ 3K1 , where P3 : v, u, w and V (3K1 ) = {x, y, z}. Theorem 3.11. Let G be a connected graph of even order n = 2k 4 with diam G = 2 that is not a star. If (a) |N ∗ (u)| + |N ∗ (v)| = 0 for every two nonadjacent vertices u and v of G, or (b) |N ∗ (u)| + |N ∗ (v)| = 1 for every two nonadjacent vertices u and v of G, then G is not uniformly cordial. Proof. First, assume that G satisfies (a). Then for vertices u, v, and w, whenever uv, vw ∈ / E(G), we have uw ∈ / E(G); for otherwise, N ∗ (u, v) = ∅, contrary to hypothesis. This implies that G is a complete k-partite graph for some k 2. Since G is not a star, it follows by Propositions 2.1 and 3.3 that G is not uniformly cordial. Next, assume that G satisfies (b). By Lemma 3.10, G = rP 3 ∪ sK 1 , where r 1 and s 0, and so n = 3r + s = 2k for some integer k 2. We consider three cases. Case 1: 3r k. Partition V (G) into two subsets V0 and V1 with |V0 | = |V1 | = k such that V (rP 3 ) ⊆ V0 . Define a k − 2r = k(k − 1) − 2r and friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = 2 2 m1 (f ) = k 2 . Since c(f ) = k + 2r 4, it follows that f is not cordial. Case 2: 3r > k and k ≡ 0 (mod3). Then k = 3t for some t 1. Partition V (G) into two subsets V0 and V1 such that |V0 | = |V1 | = kand V (tP 3 ) ⊆ V0 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then k m0 (f ) = 2 − 2r = k(k − 1) − 2r and m1 (f ) = k 2 . Since c(f ) = k + 2r 4, it follows that f is not cordial. 2
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Case 3: 3r > k and k ≡ / 0 (mod 3). Let P3 : v1 , v2 , v3 in G and let r = k/3. Partition V (G) into two subsets V0 and V1 with |V0 | = |V1 | = k such that V0 consists of V (r P3 ) and v1 , or V0 consists of V (r P3 ) and {v1 , v2 }, according to whether k ≡ 1 (mod 3) ork ≡ 2 (mod 3). Define a friendly labeling f by assigning i to every vertex in Vi for k − 2(r − 1) − 1 and m1 (f ) = k 2 − 1. Since c(f ) = k + 2r − 2 2, it follows that f is i = 0, 1. Then m0 (f ) = 2 2 not cordial. In order to determine which of the remaining connected graphs of even order and diameter 2 are uniformly cordial, we first present two lemmas. Lemma 3.12. If G = (K p + Ks ) + e, where p 3, s 1, and p + s is even, then G is not uniformly cordial. Proof. Let X = V (K p ) = {x1 , x2 , . . . , xp }, Y = {y1 , y2 , . . . , ys }, p + s = 2k 4, and e = x1 x2 . We consider two cases. Case 1: s = 1. Partition V (G) into V0 and V1 with |V0 | = |V1 | = k such that x1 , y1 ⊆ V0 and x2 ∈ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = k − 1 and m1 (f ) = k + 1. Since c(f ) = 2, it follows that f is not cordial, a contradiction. Case 2: s 2. There are two subcases. Subcase 2.1: p k + 1. Partition V (G) into V0 and V1 with |V0 | = |V1 | = k such that x1, x2 ∈ V0 ⊆ X. Define a friendly labeling f by assigning i to every vertex in Vi for i =0, 1. Then m0 (f )=1+(k−s)s + 2s =(2+2sk−s 2 −s)/2 and m1 (f ) = sk. Since c(f ) = (s 2 + s − 2)/2 2, it follows that f is not cordial, a contradiction. Subcase 2.2: p k. Partition V (G) into V0 and V1 with |V0 | = |V1 | = k such ⊆ V0 . Define a friendly thatX k p labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = 1 + 2 − and m1 (f ) = k 2 . Since 2 2 p p c(f ) = k + − 1 1 + 4, it follows that f is not cordial, a contradiction. 2 2 Lemma 3.13. Let G be a connected graph of even order n 4 with diam G = 2. If {|N ∗ (u)| + |N ∗ (v)| : u, v ∈ V (G)
and
uv ∈ / E(G)} = {0, 1},
then G = (rK 1 ∪ K2 ) + F for some integer r 1 and for some graph F for which {|NF∗ (u)| + |NF∗ (v)| : u, v ∈ V (F ) and
uv ∈ / E(F )} ⊆ {0, 1}.
Proof. Let u and v be two nonadjacent vertices of G such that |N ∗ (u)|+|N ∗ (v)|=1, say N ∗ (u, v)=∅ and N ∗ (v, u)={w} for some vertex w ∈ V (G). Necessarily, w is not adjacent to u. However, w is adjacent to every vertex in N (u, v); for otherwise, |N ∗ (u)| + |N ∗ (v)|2. Let X = {x ∈ V (G) : ux ∈ / E(G) and |N ∗ (u, x)| + |N ∗ (x, u)| = 1}, Y = {y ∈ V (G) : uy ∈ / E(G) and |N ∗ (u, y)| + |N ∗ (y, u)| = 0}. Thus v ∈ X and each vertex in Y is adjacent to every vertex in N (u, v) = N (u). We first verify the following two claims. Claim 1. X = {v, w}. Proof of Claim 1. First we show that {v, w} ⊆ X. We already observed that v ∈ X. Since N ∗ (v, u) = {w}, it follows that vw ∈ E(G) and uw ∈ / E(G). Thus v ∈ N ∗ (w, u) and so |N ∗ (u, w)| + |N ∗ (w, u)| = 1. Thus w ∈ X. Hence {v, w} ⊆ X. Next, we show that X ⊆ {v, w}. Assume, to the contrary, that X{v, w}. Then there exists x ∈ X − {v, w}. So x is not adjacent to u and |N ∗ (u, x)| + |N ∗ (x, u)| = 1. First, we show that x is adjacent to neither v nor w. Assume, without loss of generality, that x is adjacent to v. Hence w, x ∈ N ∗ (v, u), which is impossible. We now complete the proof of Claim 1 by considering two cases.
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Case 1: N ∗ (u, x) = ∅. Then N ∗ (u, x) = {z} for some z ∈ V (G). Thus z is adjacent to u and z is not adjacent to x. Consider the two nonadjacent vertices x and z. Since (1) z is adjacent to v and w and (2) x is adjacent to neither v nor w, it follows that v, w ∈ N ∗ (z, x), which is impossible. Case 2: N ∗ (x, u) = ∅. Let N ∗ (x, u) = {z }. Thus z is adjacent to x and z is not adjacent to u. If z is adjacent to v, then w, z ∈ N ∗ (v, u), a contradiction. Thus z is not adjacent to v. Similarly, z is not adjacent to w. We now consider the two nonadjacent vertices v and x. Since w ∈ N ∗ (v, x) and z ∈ N ∗ (x, v), a contradiction is produced. Therefore, X = {v, w}, which completes the proof of Claim 1. Claim 2. Y ∪ {u} is an independent set in G. Proof of Claim 2. Certainly, Claim 2 is true if Y = ∅. Thus we may assume that Y = ∅. Assume, to the contrary, Y ∪ {u} is not independent in G. Since u is not adjacent to any vertex in Y , there exist adjacent vertices y, y ∈ Y . However, then, y ∈ N ∗ (y, u), contradicting that N ∗ (y, u) = ∅. This completes the proof of Claim 2. Since vw ∈ E(G), it follows by Claim 1 that X = K2 . By Claim 2, we have Y ∪ {u} = rK 1 for some integer r 1. Let F = N (u, v). Since each vertex in X ∪Y ∪{u} is adjacent to every vertex in F , it follows that G=(rK 1 ∪K2 )+F . If u ∈ V (F ), then NF (u) = NG (u) − V (rK 1 ∪ K2 ). Hence, if u and v are two nonadjacent vertices of F , then ∗ (u) and N ∗ (v) = N ∗ (v). This implies that NF∗ (u) = NG F G ∗ ∗ |NF∗ (u)| + |NF∗ (v)| = |NG (u)| + |NG (v)|1
for every two nonadjacent vertices u and v in F .
Theorem 3.14. Let G be a connected graph of even order n 4 with diam G = 2. If {|N ∗ (u)| + |N ∗ (v)| : u, v ∈ V (G)
and uv ∈ / E(G)} = {0, 1},
(1)
then G is not uniformly cordial. Proof. Assume, to the contrary, that there exist connected uniformly cordial graphs of even order at least 4 with diameter 2 that satisfy (1). Among these graphs, let G be one of minimum order n=2k 4. By Lemma 3.13, G=(rK 1 ∪K2 )+F for some integer r 1 and some graph F for which {|NF∗ (u)| + |NF∗ (v)| : u, v ∈ V (F )
and
uv ∈ / E(F )} ⊆ {0, 1}.
(2)
We consider two cases. Case 1: F is disconnected. There are two subcases. Subcase 1.1: F is empty. Then F = tK 1 for some integer t 2 and so G = (rK 1 ∪ K2 ) + tK 1 . Let s=r+2 3, and let Ks,t be the complete bipartite graph with partite sets X={x1 , x2 , . . . , xs } and Y ={y1 , y2 , . . . , yt }. Then G is obtained from Ks,t by adding an edge joining two vertices of X, say x1 x2 . There are two subcases. Subcase 1.1.1: 3s t. Partition V (G) into V0 and V1 such that |V0 | = |V1 | = k and X ⊆ V0 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = s(k − s) + 1 = sk − s 2 + 1 and m1 (f ) = sk. Since c(f ) = s 2 − 18, it follows that f is not cordial, a contradiction. Subcase 1.1.2: 2 t < s. Partition V (G) into V0 and V1 with |V0 | = |V1 | = k such that Y ∪ {x1 } ⊆ V0 and x2 ∈ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i = 0, 1. Then m0 (f ) = t (k − t) = tk − t 2 and m1 (f ) = tk + 1. Since c(f ) = 1 + t 2 5, it follows that f is not cordial, a contradiction. Subcase 1.2: F is not empty. We claim that F =tK 1 ∪K2 for some integer t 1; otherwise, there exist two nonadjacent vertices u and v in F with |NF∗ (u)| + |NF∗ (v)|2, contradiction property (2) possessed by F . Hence, as claimed, G = (rK 1 ∪ K2 ) + (tK 1 ∪ K2 ). Let p = r + 2 3, q = t + 2 3, and let Kp,q be the complete bipartite graph with partite sets X = {x1 , x2 , . . . , xp } and Y = {y1 , y2 , . . . , yq }. We may assume that p q and that G is obtained from Ks,t by adding the two edges x1 x2 and y1 y2 . There are two subcases.
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Fig. 4. The graph G = (rK 1 ∪ K2 ) + F .
Subcase 1.2.1: p = q. Define a friendly labeling f by assigning 0 to every vertex in X and 1 to every vertex in Y . Then m0 (f ) = 2 and m1 (f ) = k 2 . Since c(f ) = k 2 − 2 7, it follows that f is not cordial, a contradiction. Subcase 1.2.2: p < q. Partition V (G) into two subsets V0 and V1 with |V0 |=|V1 |=k such that X∪{y1 } ⊆ V0 and y2 ∈ V1 . Define a friendly labeling f by assigning i to every vertex in Vi for i =0, 1. Then m0 (f )=p(k −p)+1=pk −p2 +1 and m1 (f ) = pk + 1. Since c(f ) = p 2 9, it follows that f is not cordial, a contradiction. Case 2: F is connected. There are two subcases. Subcase 2.1: F is complete. Then G = (rK 1 ∪ K2 ) + Ks = (K r+2 + Ks ) + e for some integer s 1. It follows by Lemma 3.12 that G is not uniformly cordial, a contradiction. Subcase 2.2: F is not complete. Then diam F = 2; for otherwise, there exist two vertices z1 and z2 in F such that dF (z1 , z2 )3, implying that |NF∗ (z1 )| + NF∗ (z2 )|2, contradicting property (2) of F . Let V (K2 ) = {x, y}, where K2 is the subgraph in G = (rK 1 ∪ K2 ) + F (see Fig. 4). We consider two cases. Subcase 2.1.1: F has odd order. Let V (F ) = 2p + 13 and |V (G) − V (F )| = 2q + 1 3. Since F = K3 , it follows that F is not uniformly cordial. So there is a friendly labeling f of F that is not cordial. We may assume that Ui is the set of vertices of F labeled i by f for i = 0, 1, where |U0 | = p + 1 and |U1 | = p. Partition V (G) − V (F ) into two subsets V0 and V1 with |V0 | = q and |V1 | = q + 1 such that {x, y} ⊆ V1 . Define a friendly labeling f of G by f (v) = f (v) if v ∈ V (F ) and f (v) = i if v ∈ Ui for i = 0, 1. Then m0 (f ) = m0 (f ) + q(p + 1) + p(q + 1) + 1 and m1 (f ) = m1 (f ) + qp + (q + 1)(p + 1). Since c(f ) = c(f ) 2, it follows that f is not cordial. Subcase 2.1.2: F has even order. Let |V (F )| = 2p 4 and |V (G) − V (F )| = 2q = 2k − 2p 4. Since G is a uniformly cordial graph of minimum even order n 4 with diameter 2 that satisfies (1) and the order of F is smaller than that of G, it follows that F is not uniformly cordial. Thus there is a friendly labeling f of F that is not cordial. Assume first that m0 (f ) > m1 (f ). So c(f ) = m0 (f ) − m1 (f )2. Let Ui be the set of vertices of F labeled i by f for i = 0, 1, where |U0 | = |U1 | = p. Partition V (G) − V (F ) into two subsets V0 and V1 with |V0 | = |V1 | = q such that {x, y} ⊆ V1 . Define a friendly labeling f of G by f (v) = f (v) if v ∈ V (F ) and f (v) = i if v ∈ Ui for i = 0, 1. Then m0 (f ) = m0 (f ) + pq + pq + 1 and m1 (f ) = m1 (f ) + pq + pq. Since c(f ) = m0 (f ) − m1 (f ) = m0 (f ) − m1 (f ) + 1 = c(f ) + 1 3, it follows that f is not cordial. Next, assume that m1 (f ) − m0 (f )2. Let Ui be the set of vertices of F labeled i by f for i = 0, 1, where |U0 | = |U1 | = p. Partition V (G) − V (F ) into two subsets V0 and V1 with |V0 | = |V1 | = q such that x ∈ V0 and y ∈ V1 . Define a friendly labeling f of G by f (v) = f (v) if v ∈ V (F ) and f (v) = i if v ∈ Ui for i = 0, 1. Then m0 (f ) = m0 (f ) + pq + pq and m1 (f ) = m1 (f ) + pq + pq + 1. Since c(f ) = m1 (f ) − m0 (f ) = m1 (f ) − m0 (f ) + 1 = c(f ) + 1 3, it follows that f is not cordial.
By Theorems 3.11 and 3.14, we have the following.
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Corollary 3.15. Let G be a connected graph of even order n 4 with diam G = 2 that is not a star. If |N ∗ (u)| + |N ∗ (v)| 1 for every pair u, v of nonadjacent vertices of G, then G is not uniformly cordial. By Proposition 2.1 and Corollaries 3.9 and 3.15, we have the following. Corollary 3.16. Let G be a connected graph of even order. Then G is uniformly cordial if and only if G is a star. Combining Corollaries 3.2 and 3.16, we now present a characterization of uniformly cordial graphs. Theorem 3.17. A nontrivial connected graph G of order n is uniformly cordial if and only if (1) n = 3 and G = K3 , or (2) n is even and G = K1,n−1 . Acknowledgments We are grateful to the referee whose valuable suggestions resulted in an improved paper. References [1] M. Benson, S.M. Lee, On cordialness of regular windmill graphs, Congr. Numer. 68 (1989) 49–58. [2] I. Cahit, Cordial graphs: A weaker version of graceful and harmonious graphs, Ars Combin. 23 (1987) 201–207. [3] I. Cahit, Recent results and open problems on cordial graphs, in: Contemporary Methods in Graph Theory, Bibligraphisches Inst., Mannhiem, 1990, pp. 209–230. [4] G. Chartrand, P. Zhang, Introduction to Graph Theory, McGraw-Hill, Boston, 2005. [5] G.M. Du, Cordiality of complete k-partite graphs and some special graphs, Neimenggu Shida Xuebao Ziran Kexue Hanwen Ban 2 (1997) 9–12. [6] J.A. Gallian, A dynamic survey of graph labeling, Electron. J. Combin. 5 (2000) 1–79. [7] Y.S. Ho, S.M. Lee, S.S. Shee, Cordial labelings of unicyclic graphs and generalized Petersen graphs, Congr. Numer. 68 (1989) 109–122. [8] Y.S. Ho, S.M. Lee, S.S. Shee, Cordial labelings of the Cartesian product and composition of graphs, Ars Combin. 29 (1990) 169–180. [9] W.W. Kirchherr, On the cordiality of certain specific graphs, Ars Combin. 31 (1991) 127–138. [10] W. W. Kirchherr, Algebraic approaches to cordial labeling, Graph Theory, Combinatorics, Algorithms, and Applications, Y. Alavi, et al. (Eds.) SIAM, Philadelphia, PA, 1991, pp. 294–299. [11] W.W. Kirchherr, NEPS operations on cordial graphs, Discrete Math. 115 (1993) 201–209. [12] S. Kuo, G.J. Chang, Y.H.H. Kwong, Cordial labeling of mK n , Discrete Math. 169 (1997) 1–3. [13] S.M. Lee, A. Liu, A construction of cordial graphs from smaller cordial graphs, Ars Combin. 32 (1991) 209–214. [14] Y.H. Lee, H.M. Lee, G.J. Chang, Cordial labelings of graphs, Chinese J. Math. 20 (1992) 263–273; Y.H. Lee, H.M. Lee, G.J. Chang, Cordial labelings of graphs, Chinese J. Math. 20 (1962) 270–288 [15] E. Seah, On the construction of cordial graphs, Ars. Combin. 31 (1991) 249–254. [16] S.C. Shee, The cordiality of the path-union of n copies of a graph, Discrete Math. 151 (1996) 221–229. [17] S.C. Shee, Y.S. Ho, The cordiality of the one-point union of n-copies of a graph, Discrete Math. 117 (1993) 225–243.